Question: What is the slope of the line tangent to $f(x) = -x^{2}-x+3$ at $x = 1$ ?
Answer: The slope of the tangent line is $ \lim_{\Delta x \to 0} \frac{f(x + \Delta x) - f(x)}{\Delta x}$ $ = \lim_{\Delta x \to 0} \frac{(-(x+\Delta x)^{2}-(x+\Delta x)+3) - (-x^{2}-x+3)}{\Delta x}$ $ = \lim_{\Delta x \to 0} \frac{(-(x^{2}+2x \Delta x+\Delta x^{2})-(x+\Delta x)+3) - (-x^{2}-x+3)}{\Delta x}$ $ = \lim_{\Delta x \to 0} \frac{-x^{2}-2(x \Delta x)-\Delta x^{2}-x-\Delta x+3+x^{2}+x-3}{\Delta x}$ $ = \lim_{\Delta x \to 0} \frac{-2(x \Delta x)-\Delta x^{2}-\Delta x}{\Delta x}$ $ = \lim_{\Delta x \to 0} -2x-\Delta x-1$ $ = -2x-1$ $ = (-2)(1)-1$ $ = -3$